What path could a honeybee follow,
beginning and ending at top center,
visiting every empty cell exactly once
and dripping 2 drops of honey into the last cell?

Start by heading clockwise from the cell at top center,
which has 12 honeydrops.
Eleven other cells begin with numbers of drops
that correspond to hours.

Each step consists of moving to an adjacent cell and dripping
0, 1, 2 , 3 , 4 , 5 , 6 , 7 , 8 or 9
drops into it based on how many total drops
its (1 to 6) neighboring cells, combined,
contain at that moment.

Each number of new honeydrops is the ones digit of the surrounding total.

The two top cells indicated by arrows
are the beginning and end of the path
and should each receive 2 drops,
with the central 12 included in their totals.

No need to spoilerize a text solution.
Site implementation makes that unduly onerous.

Also no need to be daunted by apparent complexity.
The path is quite constrained topologically
and luck alone probably gives a 1/10 chance of
ending correctly.

A little well-directed reckoning can help approach this
in a way that eases recovery from inevitable addition mistakes.
(Yes, this claim follows much misdirected reckoning and many mistakes.)

The following possible sequence of seven initial steps
demonstrates how the path works.

The first two steps here drip 2 drops each
because the only adjacent honey on the first step
is 12 (last digit = 2)
and on the second step is simply 2.
The seventh step drips 5 drops as
the total of its surrounding cells is 1+3+7+7+7&hairsp;= 25
(last digit = 5).

This is meant to be convenient on paper or in a text editor.
&hairsp;
Here is a template for <pre>...</pre>:

As to how I got to this answer, I first set about finding a valid path that visited every cell. The first path I found ended with 4, so then I just needed to tweak the end of the path a little bit to get it to work.

As to how I found a path in the first place

There are a few cells in the clock which are only connected to two other cells
What this means is that cell must be traversed in a fixed order (well, two orders, either forwards or backwards). This happens in 8 places, marked with an X.

Then I started thinking about how to fill the middle circle of the clock.

The only openings into the center are now at 1, 5, 7, and 11.
So I decided the best route would be one that filled in the area around 1 o'clock, did the partial loopy path from 2 to 4 o'clock,
and then, starting near 5 o'clock, circle around the inner loop, exiting at 7 o'clock.
Then you do the partial path from 8 to 10, fill in the area around 11, and then you're done.

As to how I picked specific steps in that more general path, I honestly just picked whichever direction came first if you were thinking clockwise whenever there was a choice to make. As I said, this originally led me to a path ending with the number 4, not 2, but fixing it was fairly trivial, I just changed some choices I made filling in the area around 11 o'clock.

$\begingroup$Your analysis of the maze is even better than I had hoped, thank you for taking the trouble to lay it out (spoilered even!) and all the more so for listing the path as BA...B. I've tried a few times, and still hope to be overlooking something, but I just can't get the same number as you at two places. (continued in the next two comments)$\endgroup$
– humnOct 14 '16 at 3:35

$\begingroup$At the first touch of 2, at the end of ...AABDB, I get 4+5+2 = 1 instead of 9. Fortunately, the difference seems to cancel out within a few more steps! As arduous as this maze may be, it is actually quite forgiving.$\endgroup$
– humnOct 14 '16 at 3:36

1

$\begingroup$Halfway between 10 and 11 on the perimeter, at the end of ...AAFAE, I get 9+9 = 8 instead of 9. If this turns out to be the case, slight changes there and a little later can rescue the ending. The nearby cluster of 0s is something I hoped to see and it provides an extra path-change option that might come in handy. Don't forget to add the 12 at high noon/midnight when adding up for the final 2.$\endgroup$
– humnOct 14 '16 at 3:38

1

$\begingroup$I'm fairly positive this is right now, and I made the paths clearer by taking out lines instead of using the A-F path thing$\endgroup$
– MMAdamsOct 14 '16 at 14:19

1

$\begingroup$Very nicely done and explained. I'd just started working a bit on it. Cool stuff!$\endgroup$
– Dan RussellOct 15 '16 at 2:34

So it was time to explore structures
whose traversals could feel more like journeying
than like space-filling.
The conceptual point of departure was to simply have a
hexagon made of hexagons made of hexagons,
which led to a variety of possible forms.

The purest version of the hexagonal concept,
marked 12 here,
turned out to be too repetitive to be interesting.
Version 7 felt like a discovery more than a construction,
with it’s beautifully paradoxical harmonization of
6-fold and 7-fold symmetries,
as it naturally includes many different local configurations.
Version 10 wound up with rounded corners,
to avoid unnecessarily many small-scale decisions,
and that suggested the final clock format.

Would have been descriptively clean to
begin the maze at 1 o’clock and end at 12,
but that allowed two very different overall routes,
one that was an all-too-regular zig-zag
and another much more interesting route that included
a sneaky counterclockwise inner loop
along with some zig-zags.
Both routes had the lucky feature of
forcing the hours to be visited in order.
Hah, beginning at 12 would
allow only the more interesting route.

Now to check if this is plain too easy to
solve&hairsp;—&hairsp;or,
more likely, too tedious,
especially as arithmetic mistakes could require redoing a lot of work.
A few sloppy failed test solutions were enough to realize
that it wasn't too easy and that, surprise,
most arithmetic mistakes had only local effects.
Some analysis revealed three particularly forgiving lucky features,
labeled A, B and C.

A demonstrates that 2, 3 and 4 o'clock
add a combined 0 to the running total.
B demonstrates that any sum (or mistake)
arriving at 2 or 8 o'clock
will be multiplied by 5, which turns
all even sums$\scriptsize\raise.2ex/$mistakes into 0
and prepares odd sums$\scriptsize\raise.2ex/$mistakes
to become 0 at their inevitable next doubling.
C demonstrates that the clock’s inner loop has no net effect
on the peripheral loop at 9 o'clock—whew!

More fun than anticipated to analyze and test solve,
for hours and hours,
this was ready to be posed.